我做了一个测试(Python 3.6.4, W7x64)，看看对于一个文件夹，没有子目录，哪种解决方案是最快的，以获得具有特定扩展名的文件的完整文件路径列表。

简而言之，对于这个任务，os.listdir()是最快的，比次优os.walk()快1.7倍，比pathlib快2.7倍，比os.scandir()快3.2倍，比glob快3.3倍。 请记住，当您需要递归结果时，这些结果将会改变。如果您复制/粘贴下面的一个方法，请添加.lower()，否则在搜索.ext时将找不到.ext。

import os
import pathlib
import timeit
import glob

def a():
    path = pathlib.Path().cwd()
    list_sqlite_files = [str(f) for f in path.glob("*.sqlite")]

def b(): 
    path = os.getcwd()
    list_sqlite_files = [f.path for f in os.scandir(path) if os.path.splitext(f)[1] == ".sqlite"]

def c():
    path = os.getcwd()
    list_sqlite_files = [os.path.join(path, f) for f in os.listdir(path) if f.endswith(".sqlite")]

def d():
    path = os.getcwd()
    os.chdir(path)
    list_sqlite_files = [os.path.join(path, f) for f in glob.glob("*.sqlite")]

def e():
    path = os.getcwd()
    list_sqlite_files = [os.path.join(path, f) for f in glob.glob1(str(path), "*.sqlite")]

def f():
    path = os.getcwd()
    list_sqlite_files = []
    for root, dirs, files in os.walk(path):
        for file in files:
            if file.endswith(".sqlite"):
                list_sqlite_files.append( os.path.join(root, file) )
        break



print(timeit.timeit(a, number=1000))
print(timeit.timeit(b, number=1000))
print(timeit.timeit(c, number=1000))
print(timeit.timeit(d, number=1000))
print(timeit.timeit(e, number=1000))
print(timeit.timeit(f, number=1000))

结果:

# Python 3.6.4
0.431
0.515
0.161
0.548
0.537
0.274

这样做是可行的:

>>> import os
>>> path = '/usr/share/cups/charmaps'
>>> text_files = [f for f in os.listdir(path) if f.endswith('.txt')]
>>> text_files
['euc-cn.txt', 'euc-jp.txt', 'euc-kr.txt', 'euc-tw.txt', ... 'windows-950.txt']

你可以简单地使用pathlibs glob 1:

import pathlib

list(pathlib.Path('your_directory').glob('*.txt'))

或在循环中:

for txt_file in pathlib.Path('your_directory').glob('*.txt'):
    # do something with "txt_file"

如果你想递归你可以使用。glob('**/*.txt')

1 pathlib模块被包含在python 3.4的标准库中。但是你甚至可以在旧的Python版本(即使用conda或pip)上安装该模块的反向端口:pathlib和pathlib2。

import os

path = 'mypath/path' 
files = os.listdir(path)

files_txt = [i for i in files if i.endswith('.txt')]

许多用户都回复了os。Walk回答，其中包括所有文件，还包括所有目录和子目录及其文件。

import os


def files_in_dir(path, extension=''):
    """
       Generator: yields all of the files in <path> ending with
       <extension>

       \param   path       Absolute or relative path to inspect,
       \param   extension  [optional] Only yield files matching this,

       \yield              [filenames]
    """


    for _, dirs, files in os.walk(path):
        dirs[:] = []  # do not recurse directories.
        yield from [f for f in files if f.endswith(extension)]

# Example: print all the .py files in './python'
for filename in files_in_dir('./python', '*.py'):
    print("-", filename)

或者对于一次性不需要发电机的情况:

path, ext = "./python", ext = ".py"
for _, _, dirfiles in os.walk(path):
    matches = (f for f in dirfiles if f.endswith(ext))
    break

for filename in matches:
    print("-", filename)

如果你打算为其他东西使用匹配，你可能想让它成为一个列表，而不是一个生成器表达式:

    matches = [f for f in dirfiles if f.endswith(ext)]

import os
import sys 

if len(sys.argv)==2:
    print('no params')
    sys.exit(1)

dir = sys.argv[1]
mask= sys.argv[2]

files = os.listdir(dir); 

res = filter(lambda x: x.endswith(mask), files); 

print res